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Section 3.5
Derivative of Trigonometric Functions
2 Lectures
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Dr. Abdulla Eid
College of Science
MATHS 101: Calculus I
Dr. Abdulla Eid (University of Bahrain)
Trigonometric functions
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Review of the trigonometric functions (Pre–Calculus).
2
Limits involving trigonometric functions.
3
Derivative of the basic trigonometric functions.
4
Derivative of the functions that involve trigonometric functions.
Dr. Abdulla Eid (University of Bahrain)
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3 - Derivative of the trigonometric functions
Theorem
d
(sin x ) = cos x
dx
f 0 (x )
f 0 (x )
f 0 (x )
Dr. Abdulla Eid (University of Bahrain)
Ei
Dr
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h →0
ull
a
f (x + h ) − f (x )
sin(x + h ) − sin x
= lim
h →0
h
h
sin x cos h + cos x sin h − sin x
= lim
h →0
h
sin x (cos h − 1) + cos x sin h
= lim
h →0
h
sin x (cos h − 1)
cos x sin h
= lim
+ lim
h →0
h →0
h
h
(cos h − 1)
sin h
= sin x · lim
+ cos x lim
h →0
h →0 h
h
= sin x · 0 + cos x · 1 = cos x
f 0 (x ) = lim
f 0 (x )
d
Proof: Let f (x ) = sin x
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Homework
Exercise
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Ei
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d
(cos x ) = sin x
dx
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Trigonometric functions
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Theorem
d
(tan x ) = sec2 x
dx
sin x
cos x
d
Proof: Let f (x ) = tan x =
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(denominator)derivative of numerator − (numerator)(derivative of de
(denominator)2
cos x · cos x − sin x · (− sin x )
=
(cos x )2
2
cos x + sin2 x
=
cos2 x
1
=
cos2 x
= sec2 x
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f 0 (x ) =
Dr. Abdulla Eid (University of Bahrain)
Trigonometric functions
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Homework
Exercise
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d
(cot x ) = csc2 x
dx
Dr. Abdulla Eid (University of Bahrain)
Trigonometric functions
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Theorem
d
(sec x ) = sec x tan x
dx
1
cos x
Ei
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Proof: Let f (x ) = tan x =
(denominator)derivative of numerator − (numerator)(derivative of de
(denominator)2
cos x · 0 − 1 · (− sin x )
=
(cos x )2
1 · sin x
1 · sin x
=
=
2
cos x
cos x · cos x
1
sin x
=
·
cos x cos x
= sec x tan x
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f 0 (x ) =
Dr. Abdulla Eid (University of Bahrain)
Trigonometric functions
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Homework
Exercise
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d
(csc x ) = − csc x cot x
dx
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d
Summary
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(Derivative of the Trigonometic function)
(cos x )0 = − sin x
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(sin x )0 = cos x
(tan x )0 = sec2 x
(cot x )0 = − csc x
(sec x )0 = sec x tan x
(csc x )0 = − csc x cot x
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4 - Derivative of the functions that involve the
trigonometric functions
Example
d
Differentiate y = x 2 − cos x.
Ei
Solution:
Example
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y 0 = 2x + sin x
Find y 0 for y = x 5 sin x.
Solution:
y 0 = (derivative of first)(second) + (first)(derivative of second)
y 0 = 5x 4 sin x + x 5 cos x
Dr. Abdulla Eid (University of Bahrain)
Trigonometric functions
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Example
Differentiate y = xe x + tan x + 7.
Solution:
d
y 0 = (derivative of first)(second) + (first)(derivative of second)
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4
cos x
+
Solution:
Dr
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Example
Find y 0 for y =
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y 0 = e x + xe x + sec2 x
5
tan x .
y = 4 sec x + 5 cot x
y 0 = 4 sec x tan x − 5 csc2 x
Dr. Abdulla Eid (University of Bahrain)
Trigonometric functions
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Example
Differentiate y = (sin x + cos x ) sec x.
Solution:
Example
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y 0 = sec2 x
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y = sin x sec x + cos x sec x
1
1
y = sin x ·
+ cos x ·
cos x
cos x
y = tan x + 1
Find y 0 for y = tan x cot x.
Solution:
sin x cos x
·
cos x sin x
y =1
y=
y0 = 0
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Example
Differentiate y =
sec x
1+sec x .
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Solution:
Dr
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(denominator)derivative of numerator − (numerator)(derivative of deno
(denominator)2
(1 + sec x ) sec x tan x − sec x · sec x tan x
y0 =
(1 + sec x )2
sec x tan x + sec2 x tan x − sec2 x tan x
y0 =
(1 + sec x )2
sec x tan x
y0 =
(1 + sec x )2
y0 =
Dr. Abdulla Eid (University of Bahrain)
Trigonometric functions
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Example
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π
4?
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differentiable at x =
π
4
π
4
d
For which value(s) is the function defined by
(
ax + b,
x>
f (x ) =
cos x,
x≤
Dr
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Solution: We find
f ( π4 + h ) − f ( π4 )
f (h )
0 π
f
= lim
= lim
h →0
h →0 h
4
h
We need to compute the left and right limit and we make them equal.
π−
π
1
π+
f
= sin = √
f
=a
4
4
4
2
so we have a = √12 .Now since the function is continuous, then we must
have the right limit equal the left limit and so we have
π
π
lim+ f (x ) = lim− f (x ) → √ + b = cos → b =
4
x →0
x →0
4functions
2
Dr. Abdulla Eid (University of Bahrain)
Trigonometric
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Exercise
Ei
π
6?
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differentiable at x =
π
6
π
6
d
For which value(s) is the function defined by
(
ax + b,
x>
f (x ) =
tan x,
x≤
Dr. Abdulla Eid (University of Bahrain)
Trigonometric functions
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